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lesya692 [45]
3 years ago
6

2. The net of a square pyramid and its dimensions are shown in the diagram.

Mathematics
1 answer:
vovangra [49]3 years ago
3 0

Answer:

Step-by-step explanatio............................................

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A number is being written in scientific notation. The student moves the decimal point to the left 3 places and _____.
ELEN [110]
If you are moving the decimal point to the left, the exponent is negative. If you move it to the right, the exponent is positive.
There is only one negative exponent here, and it is the first option. 
So:
A number is being written in scientific notation. The student moves the decimal point to the left 3 places and A<span>. multiplies by 10^-3.</span>
5 0
3 years ago
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Jim goes to the store and buys 3 sodas for $1.29 each and a bag of chips for 1.59. What is Jim's subtotal?​
ASHA 777 [7]

Answer:

$5.46

Step-by-step explanation:

Total = 3*1.29 + 1.59

= 5.46

7 0
3 years ago
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If twice a number is increased by 5 and this sum is multiplied by -3, the result is -57 . What is the number?
cupoosta [38]
Set it up like this, going from words on the page to letters and variables:
-3(2x+5) = -57. Twice a number is 2x, increased by 5 is 2x+5, and that is multiplied by -3 to get a total of -57. Distribute the -3 into the parenthesis to get -6x - 15 = -57. Move the 15 over by addition to get -6x = -42. Divide both sides by -6 to get that x = 7
6 0
3 years ago
PALEEZE Help Me!! The ceiling of Katie’s living room is a square that is 16 ft long on each side. To decorate for a party, she p
frez [133]

Answer:

At least 4

Step-by-step explanation:

Photo of my work and where I got the equation from

3 0
3 years ago
(4) use the method of lagrange multipliers to determine the maximum value of f(x, y) = x a y b (the a and b are two fixed positi
wlad13 [49]
Assuming f(x,y)=x^ay^b. We have Lagrangian

L(x,y,\lambda)=x^ay^b+\lambda(x+y-1)

with partial derivatives (set to 0)[/tex]

L_x=ax^{a-1}y^b+\lambda=0\implies ax^{a-1}y^b=-\lambda
L_y=bx^ay^{b-1}+\lambda=0\implies bx^ay^{b-1}=-\lambda
L_\lambda=x+y-1=0

\implies ax^{a-1}y^b=bx^ay^{b-1}\implies -bx+ay=0
x+y-1=0\implies x+y=1

Solving this system of linear equations yields x=\dfrac a{a+b} and y=\dfrac b{a+b} as the sole critical point, which in turn gives a maximum value of f\left(\dfrac a{a+b},\dfrac b{a+b}\right)=\dfrac{a^ab^b}{(a+b)^{a+b}}.
8 0
3 years ago
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